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← JEE Main 2026 · 4 Apr · Shift 2
MathematicsSingle correct+4 / -1Question 16 of 75

JEE Main 2026 · 4 Apr · Shift 2, Mathematics question 16

Let for some α∈R\alpha\in\mathbb R, f:R→Rf:\mathbb R\to\mathbb R be a function satisfying f(x+y)=f(x)+2y2+y+αxyf(x+y)=f(x)+2y^2+y+\alpha xy for all x,y∈Rx,y\in\mathbb R. If f(0)=−1f(0)=-1 and f(1)=2f(1)=2, then the value of ∑n=15(α+f(n))\displaystyle\sum_{n=1}^5\left(\alpha+f(n)\right) is:

Choose one option
Answer and worked solution

Official answer: B

Find f. Setting x=0x=0: f(y)=2y2+y−1f(y)=2y^2+y-1 (and f(1)=2f(1)=2 ✓).

Find α. Expanding f(x+y)f(x+y) gives the extra term 4xy4xy, so α=4\alpha=4.

Sum. ∑n=15(4+2n2+n−1)=20+110+15−5=140\sum_{n=1}^5(4+2n^2+n-1)=20+110+15-5=140.

(B)

Topics:Types of functionsSpecial series sums
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